Question

已知函数y=x+1,x属于[-1,2分之3],求z=x方+y方的值域.

Answer 1

z=x^2+(x+1)^2=2x^2+2x+1=2(x+1/2)^2+1/2

x∈[-1,3/2]

则当x=-1/2时,z最小,为1/2

当x=3/2时,z最大,为17/2

所以z的值域为[1/2,17/2]